Question

Rationalizati numitorul raportului :
a) 7 supra 3√21 = ?
b) 6 supra 5√10 = ?
c) 2 supra 1+√3 = ?
d) √6 supra 5-2√6 = ?
e) 2-√5 supra 2+√5 = ?

Answer 1

$a) \frac{7}{3 \sqrt{21} } = \frac{7 \sqrt{21} }{3 \times 21} = \frac{7 \sqrt{21} }{63} = \frac{ \sqrt{21} }{9}$

$b) \frac{6}{5 \sqrt{10} } = \frac{6 \sqrt{10} }{5 \times 10} = \frac{6 \sqrt{10} }{50} = \frac{3 \sqrt{10} }{25}$

$c) \frac{2}{1 + \sqrt{3} } = \frac{2 - 2 \sqrt{3} }{1 - { \sqrt{3} }^{2} } = \frac{2(1 - \sqrt{3}) }{1 - { \sqrt{3} }^{2} } = \frac{2(1 - \sqrt{3})}{1 - 3} = \frac{2(1 - \sqrt{3}) }{ - 2} = - \frac{2(1 - \sqrt{3}) }{2} = - (1 - \sqrt{3}) = - 1 + \sqrt{3}$

$d) \frac{ \sqrt{6} }{5 - 2 \sqrt{6} } = \frac{5 \sqrt{6} + 12 }{ {5}^{2} - {(2 \sqrt{6}) }^{2} } = \frac{5 \sqrt{6} + 12 }{25 - {2}^{2} { \sqrt{6} }^{2} } = \frac{5 \sqrt{6} + 12 }{4 \times 6} = \frac{5 \sqrt{6} + 12 }{25 - 24} = \frac{5 \sqrt{6} + 12 }{1} = 5 \sqrt{6} + 12$

$e) \frac{2 - \sqrt{5} }{2 + \sqrt{5} } = \frac{4 - 2 \sqrt{5} - 2 \sqrt{5} + 5 }{ {2}^{2} - { \sqrt{5} }^{2} } = \frac{(4 + 5) + ( - 2 \sqrt{5} - 2 \sqrt{5}) }{ {2}^{2} - { \sqrt{5} }^{2} } = \frac{9 - 4 \sqrt{5} }{ {2}^{2} - { \sqrt{5} }^{2} } = \frac{9 - 4 \sqrt{5} }{4 - 5} = \frac{9 - 4 \sqrt{5} }{ - 1} = - (9 - 4 \sqrt{5}) = - 9 + 4 \sqrt{5}$